Yes. In such a case, manipulate the problem so that you get the indeterminate form 0/0 or infinity/infinity, then proceed with l'Hopital's Rule.
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Their is no # that can be multiplied into 0 other than its self, therfore it is no defined value and is called an indeterminate form.
It is indeterminate as it cannot be sure if it is 0 or 1.
because you can't divide a number by zero. it's impossible (you can't take 7 marbles and divide them into 0 groups), so the answer cannot be determined (indeterminate).
Firstly, infinity is not a number (at least in lower level mathematics). You must instead use the language of limits to describe infinity. Using limits, a function which diverges to infinity multiplied by a function which diverges to infinity has a product which also diverges to infinity. However, taking this product, and subtracting away a function which diverges to infinity is "of indeterminate form". It might converge to zero, it might be diverge to positive infinity, it might diverge to negative infinity, or it might converge to a constant. In order to figure out which one of these possibilities applies, you must get the indeterminate form into the form infinity divided by infinity or 0/0 and then apply L'Hospital's rule. Edit: Just a pet peeve of mine. It's L'Hôpital, not L'Hospital. Even textbooks don't spell it right.
3Y*12X = 0 is equivalent to YX = 0 In functional form, it is Y = z*(X = 0) where z can have any value. Here (X = 0) is 1 if X = 0 and the statement is TRUE 0 if X ≠ 0 and the statement is FALSE.